Can someone help me with this problem please. ( I need it as fast as possible).
1 answer:
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Answer:
volume of the solid generated when region R is revolved about the x-axis is π ₀∫^a ( x + b )² dx
Step-by-step explanation:
Given the data in the question and as illustrated in the image below;
R is in the region first quadrant with vertices; 0(0,0), A(a,0) and B(0,b)
from the image;
the equation of AB will be;
y-b / b-0 = x-0 / 0-a
(y-b)(0-a) = (b-0)(x-0)
0 - ay -0 + ba = bx - 0 - 0 + 0
-ay + ba = bx
ay = -bx + ba
divide through by a
y = x + ba/a
y = x + b
so R is bounded by y = x + b and y =0, 0 ≤ x ≤ a
The volume of the solid revolving R about x axis is;
dv = Area × thickness
= π( Radius)² dx
= π ( x + b )² dx
V = π ₀∫^a ( x + b )² dx
Therefore, volume of the solid generated when region R is revolved about the x-axis is π ₀∫^a ( x + b )² dx
Martin will have to mark 0.125 on the
Let the original weight of the scale be 1kg. Hence we can say the original mark for 1kg weight is 1 that is:
1kg = 1 ................ 1
Since we are to calculate where to mark for a 1/8 kg weight, we can write;
1/8 kg = x ..................2
Divide both expressions
This means Martin will have to mark 0.125 on the scale.
Learn more here: https://brainly.lat/tarea/13002863
Answer:
cross multiplication
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